Another Time Dilation Thought Experiment

[Mowgli]

Assume a stationary observer, sending a series of "pings" at the interval of one nanosecond toward a spaceship. The spaceship also has an identical pinging device, sending one ping a nanosecond. Furthermore, the spaceship has a transponder; every time it gets a ping from the observer, it sends back a reply. (Think of the ping as a laser flash and the reply as a reflection.)

 

The observer receives two serieses

1. the spaceship pings, and

2. the replies.

The observed interval between the spaceship pings is 1+d nanoseconds (d being the time dilation factor), and the interval between the replies is one nanosecond.

 

As far as the spaceship is concerned, the observer is moving. So the pings it receives from the observer comes in at a rate of one per 1+d nanoseconds, and the spaceship is sending replies at the same rate. The spaceship is sending out its own pings one every nanosecond. Assume that one reply and one ping are sent out at an instant t = 0. They will be observed together in oberver's time, let's say at the instant t_o = 0. The next ping is sent out at t = 1 nanosecond (observed at t_o = 1+d ns), and the next reply is sent out at t=1+d nanosenconds (observed at t_o = 1ns).

 

In other words, the spaceship ping sent out at t = 1 nanosecond is observed later than the reply sent at t = 1+d nanoseconds. Why is the ping sent at an earlier time received later by the observer? Where do they cross paths?

 

(In order to minimize the light travel time effect, let's assume that the spaceship is flying over the observer's head, and at the instant t=0.5ns, it is at its point of closest approach to the observer.)

 

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http://www.TheUnrealUniverse.com

[swansont]

Assume that one reply and one ping are sent out at an instant t = 0. They will be observed together in oberver's time' date=' let's say at the instant t_o = 0. The next ping is sent out at t = 1 nanosecond (observed at t_o = 1+d ns), and the next reply is sent out at t=1+d nanosenconds (observed at t_o = 1ns).

[/quote']

 

Why will the observation be at t_o = 1 ns? I think you aren't switching frames properly.

[Mowgli]

Why will the observation be at t_o = 1 ns? I think you aren't switching frames properly.

Because the observer is sending out his pings at the rate of one every nanosecond, he expects to get the replies every nanosecond (after correcting for the light travel time effects).

[Klaynos]

But he wouldn't get the replies ever 1ns, because, modeling the pings as individule photons

 

One photon leaves the observers at t=0, hit the mirror

 

|______ ~~> ______|

|______ <~~ ______|

|<---------s------->|

 

And hits back at t=a, has traveled a distance of s

 

The second photon launched at t=1

|________~~>________|

|________<~~________|

|<---------r--------- ->|

 

This has traveled distance r so gets back at time t= 1 + a +d

d is some variable, and can probably be modeld as t= (1+a)g

[swansont]

Because the observer is sending out his pings at the rate of one every nanosecond, he expects to get the replies every nanosecond (after correcting for the light travel time effects).

 

I think you're using the wrong frame to measure the nanosecond. It's a nanosecond by his clock, not by the other clock. You do have to account for light travel; you can't just ignore it, and you need more rigor (t+d is bad form, since dilation is a multiplicative, not additive, factor) Relativity invites enough confusion on its own, without using sloppy notation.

[Mowgli]

you need more rigor (t+d is bad form, since dilation is a multiplicative, not additive, factor) Relativity invites enough confusion on its own, without using sloppy notation.

Sorry. In fact, it was t(1+d), which is multiplicative. t just happened to be 1 ns.

I think you're using the wrong frame to measure the nanosecond. It's a nanosecond by his clock, not by the other clock.

Hmm... I have to think about this.

 

As you surely see, I'm basically doing the twin paradox here, trying to eliminate the need to accelerate or decelerate one of the twins.

You do have to account for light travel; you can't just ignore it.

Yes, I was saying "after correcting for light travel time effects", not ignoring them.

 

Along that line, can I ask you something else? If the observer is a point mass, and the spaceship is in a stationary orbit around the observer, then can I say that the spaceship is an intertial frame and apply SR?

[Mowgli]

But he wouldn't get the replies ever 1ns' date=' because, modeling the pings as individule photons

 

One photon leaves the observers at t=0, hit the mirror

 

|______ ~~> ______|

|______ <~~ ______|

|<---------s------->|

 

And hits back at t=a, has traveled a distance of s

 

The second photon launched at t=1

|________~~>________|

|________<~~________|

|<---------r--------- ->|

 

This has traveled distance r so gets back at time t= 1 + a +d

d is some variable, and can probably be modeld as t= (1+a)g[/quote']

Yes, the fact that the spaceship is at a different position by the time light traverses the distance to/from the observer is what I meant by the light travel time effect. I tried to minimize this effect by saying that the spaceship is traveling perpendicular to the observer.

 

As swansont rightly pointed out, I still shouldn't ignore the travel time effect. Which is why I was wondering if I could keep the spaceship traveling perpendicular to the observer all the time, leading to my question, "If the observer is a point mass, and the spaceship is in a stable orbit around the observer, then can I say that the spaceship is an intertial frame and apply SR?" (I said "stationary" instead of "stable" in the previous post; it was a mistake)

[swansont]

As you surely see' date=' I'm basically doing the twin paradox here, trying to eliminate the need to accelerate or decelerate one of the twins.

[/quote']

 

Right, so one knows what the answer must be, and if you get a wrong answer, there is a problem with the method of getting the solution.

 

Along that line, can I ask you something else? If the observer is a point mass, and the spaceship is in a stationary orbit around the observer, then can I say that the spaceship is an intertial frame and apply SR?

 

 

It depends on what you are trying to do. If a single interaction with the spaceship occurs and is essentially instantaneous, then its motion during that instant is in a straight line, and SR should be fine. But over longer periods the non-inertial frame will introduce effects, and you must account for this. (the Sagnac effect, typically)

[Mowgli]

Right, so one knows what the answer must be, and if you get a wrong answer, there is a problem with the method of getting the solution.

Spoken like a true believer

And sounds like Einstein himself, "If the facts don't fit the theory, change the facts."

[Klaynos]

They're not facts without experimental evidence...

 

But if a thought experiment contradicts experimentally "proven" theory, then the thought experiment is probably wrong...

[swansont]

Spoken like a true believer

And sounds like Einstein himself' date=' "If the facts don't fit the theory, change the facts."[/quote']

 

Given the choice between a "thought experiment" lacking a rigorous analysis, and the extensive experimental confirmation of SR, I know on which horse I am going to place my bet.

[Mowgli]

But if a thought experiment contradicts experimentally "proven" theory' date=' then the thought experiment is probably wrong...[/quote']

This is another invitation to quote the great master again

Einstein: "No amount of experimentation can ever prove me right; a single experiment can prove me wrong."

I like to believe that he meant thought experiments as well (though not the ones lacking rigor.)

Given the choice between a "thought experiment" lacking a rigorous analysis' date=' and the extensive experimental confirmation of SR, I know on which horse I am going to place my bet.

[/quote']

Fair enough. Let me try to take out the lack of rigor coming from the light travel time effects.

 

Let the observer and the spaceship be in the same orbit around a point mass. Let them be on diametrically opposite sides. The orbital velocity is v. Since both of them are in orbit, they are intertial frames and SR applies - by Einstein's Equivalence Principle.

 

The relative velocity between the observer and the spaceship is u (= 2v/sqrt(1+v^2/c^2), though it doesn't matter what u is as long as it is not zero). The consequent time dilation factor is 1+d.

 

Now the rest of my original post works:

 

Assume the observer is sending a series of "pings" at the interval of one nanosecond toward the spaceship. The spaceship also has an identical pinging device, sending one ping a nanosecond. Furthermore, the spaceship has a transponder; every time it gets a ping from the observer, it sends back a reply. (Think of the ping as a laser flash and the reply as a reflection.)

 

The observer receives two serieses

1. the spaceship pings, and

2. the replies.

The observed interval between the spaceship pings is 1+d nanoseconds (d being the time dilation factor), and the interval between the replies is one nanosecond.

 

As far as the spaceship is concerned, the observer is moving. So the pings it receives from the observer comes in at a rate of one per 1+d nanoseconds, and the spaceship is sending replies at the same rate. The spaceship is sending out its own pings one every nanosecond. Assume that one reply and one ping are sent out at an instant t = 0. They will be observed together in oberver's time, let's say at the instant t_o = 0. The next ping is sent out at t = 1 nanosecond (observed at t_o = 1+d ns), and the next reply is sent out at t=1+d nanosenconds (observed at t_o = 1ns).

 

In other words, the spaceship ping sent out at t = 1 nanosecond is observed later than the reply sent at t = 1+d nanoseconds. Why is the ping sent at an earlier time received later by the observer?

[swansont]

This is another invitation to quote the great master again

Einstein: "No amount of experimentation can ever prove me right; a single experiment can prove me wrong."

I like to believe that he meant thought experiments as well (though not the ones lacking rigor.)

 

 

Klaynos put proven in quotes, indicating a good understanding that scientific inquiries do not actually prove in the sense of mathematical proofs, i.e. they are not deductive. However, something that has been confirmed to a high degree, and is usually accepted as fact. As Stephen J Gould explained, "In science, fact can only mean confirmed to such a degree that it would be perverse to withhold provisional assent. I suppose that apples might start to rise tomorrow, but the possibility does not merit equal time in physics classrooms." IOW, it is reasonable to consider gravity (insofar as Newtonian gravity applies) to be a proven fact.

 

AFAIK SR is self-consistent, so a thought experiment, using SR, isn't going to disprove it.

[Mowgli]

AFAIK SR is self-consistent' date=' so a thought experiment, using SR, isn't going to disprove it.[/quote']

SR is also based on a few assumptions -- the homogeneity of space and time is one of them, from which it follows that you need to derive the coordinate transformation only for receding frames of reference. One may be able to expose the flaw in this assumption (if there is one) using some thought experiment. At least, one can hope...

[swansont]

SR is also based on a few assumptions -- the homogeneity of space and time is one of them, from which it follows that you need to derive the coordinate transformation only for receding frames of reference. One may be able to expose the flaw in this assumption (if there is one) using some thought experiment. At least, one can hope...

 

All you can hope to do is describe an experiment and what you results you would get if the experiment were actualy done. i.e. you can make a prediction, based on some assumption. But I don't see that what you've done here; if I get a more than few minutes free later I'll try and work through it.

[timo]

Mowgli, I don´t really understand what point you´re trying to make but I think the setup would be much clearer to you if you drew a spacetime-diagram of the system. I´ve done some sketchy version of it from what I understood in your 1st post (see image below).

 

Note that I have tried to keep the distances between the intervals equal but that neither the gamma nor the actual coordinates are calculated out. My proposal to you would be drawing a similar diagram according to your proposed setup and calculate the coordinates of the interesting events. Then, you can transfer them to another frame of reference (like that of the spaceship) using the normal lorentz matrices and see what the diagram looks like in that frame. I assume that would clear most of your questions/irriations.

[Mowgli]

Mowgli' date=' I don´t really understand what point you´re trying to make but I think the setup would be much clearer to you if you drew a spacetime-diagram of the system. I´ve done some sketchy version of it from what I understood in your 1st post (see image below).

[/quote']

Thanks Atheist for taking the time to draw the diagram. I will follow your suggestion and work on a similar diagram and the associated transformation. Will post it tomorrow.

[Mowgli]

A quick question: two satellites in the same orbit around a gravitational body - are they supposed to have zero relative velocity? What if they are at diametrically opposite points, so that one is traveling at a velocity v and the other at -v, where v is the orbital speed?

[swansont]

A quick question: two satellites in the same orbit around a gravitational body - are they supposed to have zero relative velocity? What if they are at diametrically opposite points, so that one is traveling at a velocity v and the other at -v, where v is the orbital speed?

 

They are stationary in a rotating coordinate system, so I think all you have to do is account for the Sagnac effect.

[Mowgli]

So if there was another earth on the other side of the sun, there wouldn't be any SR effects between our earth and the other one, right? Thanks.

[CPL.Luke]

but like was said before, those aren't inertial frames and you have to account for some other effects.

 

 

Also leave the quoting to the social sciences, were physisists here show us the numbers

[Mowgli]

Correct me if I'm wrong, but I thought a frame in free fall in a gravitational field is an inertial frame where SR applies. Of course, I understand that I have to account for the Sagnac effect, but there are no GR effects.

[pablo d]

Let the observer and the spaceship be in the same orbit around a point mass. Let them be on diametrically opposite sides. The orbital velocity is v. Since both of them are in orbit, they are intertial frames and SR applies - by Einstein's Equivalence Principle.

 

If they are in orbit around something, doesn't that introduce acceleration and thus wouldn't GR apply instead?

[Mowgli]

If they are in orbit around something, doesn't that introduce acceleration and thus wouldn't GR apply instead?

When something is in an orbit, gravity and acceleration kind of cancel each other - hence weightlessness. I think the equivalence principle says that such a frame is in fact an inertial frame so that SR applies.

[CPL.Luke]

no, gravity and acceleration cannot cancel eachother, I know that in GR everyhthing is generalised to make the frame in a gravitational field inertial, but SR cannot do this.